SECTION 4: TRANSMISSION LINES. ESE 470 Energy Distribution Systems
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1 SECTION 4: TRANSMISSION LINES ESE 470 Energy Distribution Systems
2 2 Introduction
3 Transmission Lines 3 Transmission and distribution of electrical power occurs over metal cables Overhead AC or DC Underground AC or DC In the U.S. nearly all transmission makes use of overhead AC lines These cables are good, but not perfect, conductors Series impedance Shunt admittance In this section of notes we ll look at how these are accounted for in equivalent circuit models
4 Electrical Properties of Transmission Lines 4 Series resistance Voltage drop (IIII) and real power loss (II 2 RR) along the line Due to finite conductivity of the line Series inductance Series voltage drop, no real power loss Only self inductance (no mutual inductance) in balanced systems Shunt conductance Real power loss (VV 2 GG) Leakage current due to corona effects or leakage at insulators Typically neglected for overhead lines Shunt capacitance Capacitance to other conductors and to ground Line-charging currents
5 5 Conductors
6 Conductors 6 Before getting into transmission line models, we ll take a look at the conductors themselves Aluminum is the most common conductor Good conductivity Light weight Low cost Plentiful supply Most common cable type combines aluminum and steel Aluminum-conductor steel-reinforced (ACSR) Bare, stranded cable Core of steel strands provides strength Outer aluminum strands provide good conductivity
7 ACSR Cables 7 ACSR cables vary based on number of aluminum conductor strands and number of steel reinforcement strands ACSR variants assigned bird code names, e.g.: Dove: 26/7 Al/Steel stranding Bluebird: 84/19 Al/Steel stranding source: Glover, Sarma, Overbye Another increasingly popular cable type is all-aluminum-alloy conductor (AAAC) Stronger Lighter Higher conductivity More expensive
8 Cables 8 Cables are sized to provide the required current-carrying capability or ampacity Number of individual strands Diameter of individual strands Strand and cable diameter commonly measured in mils 1 mmmmmm = 0.001" Cross-sectional area often measured in circular mils or cmil Area of a circle with a diameter of dd = 1 mmmmmm = cccccccc = ππ = ssss iiii Area in cccccccc of a cable with diameter dd mmmmmm: AA = dd 2
9 ACSR Cable 9 Consider, for example, Falcon ACSR cable 54/19: 54 Al strands with a core of 19 steel strands Al strand diameter: 172 mmmmmm Al strand area: 172 mmmmmm 2 = kkkkkkkkkk Steel strand diameter: 103 mmmmmm Steel strand area: 103 mmmmmm 2 = kkkkkkkkkk Cable diameter: Cable area: 1545 mmmmmm 2 = 2387 kkkkkkkkkk Ampacity: 1380 AA Weight: 10,777 llll/mmmm
10 Bundling 10 In addition to increasing cable cross-sectional area, ampacity can be increased by adding additional cables to each phase bundling Two-, three-, and four-cable bundles are common:
11 Bundling 11 Typical bundling: 345 kv: two conductors 500 kv: three conductors 765 kv: four conductors Advantages of bundling: Lower resistance Lower reactance (inductance) Increased ampacity Reduced electric field gradient surrounding phase conductor Reduced corona Reduced loss, noise, and RF interference Improved heat dissipation
12 Insulators 12 Cables are supported by towers Must connect, while retaining electrical isolation Connections are typically made through ceramic or glass insulators High-voltage lines suspended by strings of insulator discs One or two strings Two prevents sway Number of discs dictated by line voltage, e.g.: 4-6 for 69 kv for 765 kv
13 Transposition 13 Transmission-line inductance and capacitance determined by geometry Cable size and relative spacing Consider three phases laid out side-by-side Phases a and c will have similar inductance and capacitance Inductance and capacitance of phase b will differ
14 Transposition 14 Transposition Switch the position of each phase twice along the length of the line Each phase occupies each position for one third of the line length Line remains balanced
15 15 Medium- and Short-Line Models
16 Short-Line Model 16 How we choose to model the electrical characteristics of a transmission line depends on the length of the line Short-line model: < ~80 kkkk Lumped model Account only for series impedance Neglect shunt capacitance RR and ωωωω are resistance and reactance per unit length, respectively Each with units of Ω/mm ll is the length of the line
17 Medium-Line Model 17 Medium-line model nominal-ππ model: 80 kkkk < ll < 250 kkkk Lumped model Now include shunt capacitance Still a lumped model zz = RR + jjjjjj Ω/mm and ZZ = zzzz Ω yy = ωωωω SS/mm and YY = yyyy SS All impedances and admittances lumped into one or two circuit components
18 Transmission Lines as Two-Port Networks 18 Before moving on to a model for longer transmission lines, we ll look at an alternative tool for characterizing transmission line networks We can treat transmission lines as general two-port networks As two-port networks, we can characterize transmission lines with their ABCD parameters or chain parameters
19 ABCD Parameters 19 ABCD (or chain or transmission or cascade) parameters define the following two-port relationships VV 1 = AAVV 2 + BBII 2 II 1 = CCVV 2 + DDII 2 In matrix form, the chain-parameter equations are VV 1 II 1 = AA BB CC DD AA, BB, CC, and DD are, in general, complex numbers AA and DD are dimensionless BB is an impedance with units of Ω CC is an admittance with units of SS VV 2 II 2 VV 1 and VV 2 are line-to-neutral voltages If the network is reciprocal, then AAAA BBBB = 1 If the network is symmetric, then AA = DD
20 ABCD Parameters Short-Line Model 20 We ll now derive the ABCD parameters for the shorttransmission-line model Applying KVL around the loop gives our first equation VV SS II RR ZZ VV RR = 0 So, VV SS = VV RR + ZZII RR AA = 1 and BB = ZZ
21 ABCD Parameters Short-Line Model 21 Applying KCL gives the second equation II ss = II RR and CC = 0 and DD = 1 The short-line ABCD matrix is AAAAAAAA = 1 ZZ 0 1 Note that, due to symmetry and reciprocity, AA = DD and AAAA BBBB = 1
22 ABCD Parameters Medium-Line Model 22 Next, for the medium-transmission-line model Applying KVL around the loop gives our first equation VV SS II RR + VV RR YY 2 ZZ VV RR = 0 VV SS = 1 + YYZZ 2 VV RR + ZZII RR This is the first chain parameter equation, where AA = 1 + YYYY 2 and BB = ZZ
23 ABCD Parameters Medium-Line Model 23 For the second equation, apply KCL at the sending end II ss VV ss YY 2 II RR VV RR YY 2 = 0 Substituting in our previous expression for VV SS II SS = VV RR YY 2 + II RR YYYY 2 II SS = 2 + YYYY 2 YY 2 VV RR + YYYY 2 II RR YY 2 VV RR YYYY 2 II RR This is the second chain-parameter equation, where CC = 1 + YYYY 4 YY and DD = 1 + YYYY 2
24 ABCD Parameters Medium-Line Model 24 The medium-line chain parameters are AAAAAAAA = 1 + YYZZ 2 ZZ 1 + YYYY 4 YY 1 + YYYY 2 Again, note that, due to symmetry and reciprocity, AA = DD and AAAA BBBB = 1 Also note that allowing YY 0 yields the chain parameters for the short-line model
25 Cascading Two-Port Networks 25 ABCD parameters or chain parameters are also called cascade parameters If we cascade multiple two-port networks, the ABCD parameter matrix for the cascade is the product of the individual ABCD parameter matrices AAAAAAAA = AA 1AA 2 + BB 1 CC 2 AA 1 BB 2 + BB 1 DD 2 CC 1 AA 2 + DD 1 CC 2 CC 1 BB 2 + DD 1 DD 2
26 Cascaded Two-Ports - Example 26 For example, consider the cascade of the following two two-port networks ABCD parameters for the first network are AAAAAADD 1 = And for the second network So the overall ABCD matrix is 1 + YY 1ZZ YY 1ZZ 1 4 AAAAAADD 2 = 1 ZZ 0 1 = 1 4 Ω 0 1 AAAAAAAA = ZZ 1 YY YY 1ZZ jjj 6 + jjjj Ω 4 + jjj SS 15 + jjjj = 1 + jjj 2 Ω 4 + jjj SS 1 + jjj
27 Cascaded Two-Ports - Example 27 If a sending-end voltage of VV SS = VV is applied, and no load is connected, what is the receiving-end voltage? VV SS = VV and II RR = 0 AA VV SS = AAVV RR + BBII RR = 1 + jjj VV RR The no-load receiving-end voltage is VV RR = jjj = 7.06 jjjj.2 VV VV RR = VV
28 Voltage Regulation 28 The voltage at the receiving end of a line will change depending on the load placed on the line Magnitude of this change is quantified as voltage regulation Voltage regulation: Change in receiving-end voltage from no load to full load, expressed as a percentage of the full-load voltage %VVVV = VV RRRRRR VV RRRRRR 100% VV RRRRRR Typically, transmission lines are designed to limit voltage regulation to about 10% As we ve seen, the no-load voltage is given by VV RRRRRR = VV SS AA
29 Voltage Regulation Example Consider a three-phase, 60 Hz, 345 kv transmission line with the following properties 200 km long zz = jjj.35 Ω/kkkk, yy = jjj.2 μμμμ/kkkk Full load is 700 MW at 95% of the rated voltage and a power factor of 0.99 leading Determine: ABCD parameters for an appropriate transmission-line model Phase shift between sending- and receiving-end voltages at full load Percent voltage regulation
30 Voltage Regulation Example Line is 200 km long, so a nominal-ππ model is appropriate where The ABCD parameters are ZZ = zz 200 kkkk = jjjj Ω YY = yy 200 kkkk = jjjjj μμμμ AAAAAAAA = 1 + YYZZ 2 ZZ 1 + YYYY 4 YY 1 + YYYY 2 = jjj jjjj Ω jjjjj μμμμ jjj.0027 AAAAAAAA = Ω μμμμ
31 Voltage Regulation Example At full load the line-to-line receiving-end voltage is VV RRRRRR = 345 kkkk 0.95 = kkvv LLLL And the line-to-neutral voltage is VV RRRRRR = kkvv LLLL 3 = kkvv LLLL Using the receiving-end voltage as the reference, the receiving-end voltage phasor is VV RR = kkkk We know that the load power is given by PP = 3VV RRRRRR II RR cos θθ θθ is the power-factor angle (leading, so it s negative) θθ = cos 1 pp. ff. = cos = 8.1
32 Voltage Regulation Example The receiving-end current phasor is II RR = PP 3VV RRRRRR cos θθ θθ = II RR = kkkk 700 MMMM kkkk To determine the phase shift from sending to receiving end, use chain parameters to determine VV SS (line-to-neutral) VV SS = AAVV RR + BBII RR VV SS = kkkk Ω kkkk VV SS = kkvv LN So, the phase shift along the line is 26.1
33 Voltage Regulation Example The percent voltage regulation is given by %VVVV = VV RRRRRR VV RRRRRR 100% VV RRRRRR The line-to-neutral no-load voltage is VV RRRRRR = VV SS AA = = kkkk The full-load line-to-neutral voltage was given to be VV RRRRRR = kkkk So, the percent voltage regulation is %VVVV = kkkk kkkk 100% = 8.7% kkkk
34 34 Exact Transmission-Line Equations
35 Distributed Transmission Line Model 35 The medium- and short-line models are lumped models All series impedance lumped into one element Shunt admittances lumped into two elements Real lines are distributed networks Lumped models are inaccurate for long lines To treat a line as a distributed network, consider the impedance and admittance of a segment of differential length, Δxx
36 Transmission Line Differential Equations 36 Apply KVL around the differential length of line VV xx + Δxx = VV xx + II xx zzδxx VV xx+δxx VV xx Δxx = zzii xx (1) If we let the length of the line segment, Δxx, go to zero, we get ddvv xx dddd = zzii xx (2) A first-order differential equation This is a second-order segment, so we need a second first-order differential equation to describe it completely Apply KCL at xx + Δxx II xx + Δxx = II xx + VV xx + Δxx yyδxx II xx+δxx II xx Δxx = yyvv xx + Δxx (3)
37 Transmission Line Differential Equations 37 Again, letting Δxx 0 ddii xx dddd = yyvv xx (4) Our goal is a single differential equation in VV xx to describe the segment of transmission line Must eliminate II xx Solving (2) for II xx and differentiating gives ddii xx dddd = 1 ZZ dd 2 VV xx ddxx 2 (5) Substituting (5) into (4) yields the single second-order differential equation for the line segment dd 2 VV xx ddxx 2 zzzzvv xx = 0 (6)
38 Transmission Line Differential Equations 38 dd 2 VV xx ddxx 2 zzzzvv xx = 0 (6) This is a second-order, homogeneous, linear, constant-coefficient, ordinary differential equation Its characteristic equation is ss 2 zzzz = 0 The roots of the characteristic polynomial are ss = ± zzzz = ±γγ where γγ = zzzz is the propagation constant, with units of mm 1 (or rrrrrr/mm) The solution to (6) is VV xx = KK 1 ee γγγγ + KK 2 ee γγγγ (7) where KK 1 and KK 2 are unknown constants to be determined through application of boundary conditions
39 Transmission Line Differential Equations 39 We can get an expression for current by differentiating (7) and substituting back into (2) Solving for II xx dddd xx dddd = γγkk 1ee γγγγ γγkk 2 ee γγγγ = zzii xx II xx = KK 1ee γγγγ KK 2 ee γγγγ zz γγ (8) The term in the denominator of (8) is the characteristic impedance of the line, ZZ cc, with units of ohms (Ω) ZZ cc = zz γγ = zz zzzz = zz yy (9)
40 Transmission Line Differential Equations 40 Using (9), (8) becomes II xx = KK 1ee γγγγ KK 2 ee γγγγ ZZ cc (10) We can now apply boundary conditions to determine the two unknown coefficients, KK 1 and KK 2 At the receiving end of the line, which we ll define to be xx = 0, we have So, VV 0 = VV RR and II 0 = II RR VV 0 = KK 1 + KK 2 = VV RR II 0 = KK 1 KK 2 ZZ cc = II RR
41 Transmission Line Differential Equations 41 Solving each equation for KK 2 KK 2 = VV RR KK 1 = KK 1 ZZ cc II RR Solving for KK 1, then back-substituting to solve for KK 2 gives KK 1 = VV RR+ZZ cc II RR 2 KK 2 = VV RR ZZ cc II RR 2 Substituting into (7) and (10) VV xx = II xx = VV RR+ZZ cc II RR 2 VV RR+ZZ cc II RR 2ZZ cc ee γγγγ + VV RR ZZ cc II RR 2 ee γγγγ (11) ee γγγγ VV RR ZZ cc II RR 2ZZ cc ee γγγγ (12)
42 Transmission Line Differential Equations 42 Collecting VV RR and II RR terms in (11) and (12) VV xx = eeγγγγ +ee γγγγ 2 VV RR + ZZ cc ee γγγγ ee γγγγ 2 II RR (13) II xx = 1 ZZ cc ee γγγγ ee γγγγ 2 VV RR + eeγγγγ +ee γγγγ 2 II RR (14) The terms in parentheses can be represented as hyperbolic functions VV xx = cosh γγγγ VV RR + ZZ cc sinh γγγγ II RR (15) II xx = 1 ZZ cc sinh γγγγ VV RR + cosh γγγγ II RR (16)
43 Transmission Line Differential Equations 43 Equations (15) and (16) give the chain parameters for the two-port network between a point at location xx along the line and the receiving end AAAAAAAA xx = cosh γγγγ 1 sinh γγγγ ZZ cc ZZ cc sinh γγγγ cosh γγγγ For chain parameters between sending and receiving ends, we set xx = ll AAAAAAAA = cosh γγll 1 sinh γγll ZZ cc ZZ cc sinh γγll cosh γγll
44 Propagation Constant 44 We defined the propagation constant as γγ = zzzz This is, in general, a complex value γγ = αα + jjjj (17) The real part, αα, is the attenuation constant Represents loss along the line Due to series resistance and/or shunt conductance The imaginary part, ββ, is the phase constant Represents change in phase along the line Due to series reactance and/or shunt susceptance
45 Long-Line Equivalent ππ Circuit 45 Now that we have exact ABCD parameters for a distributed transmission line, we can create an equivalent ππ circuit Here we re using ZZZ and YYY to distinguish from ZZ = zzzz and YY = yyyy of the lumped, nominal ππ-circuit model Equating the ABCD parameters with those for the equivalent ππ circuit above cosh γγγγ 1 sinh γγγγ ZZ cc ZZ cc sinh γγγγ cosh γγγγ = 1 + YY ZZ 2 YY 1 + YY ZZ 4 ZZZ 1 + YY ZZ 2
46 Long-Line Equivalent ππ Circuit 46 Equating the BB parameters, we see that ZZ = ZZ cc sinh γγγγ (18) Using (18) in the AA-parameter equation gives 1 + YY 2 ZZ cc sinh γγγγ = cosh γγγγ YY 2 = cosh γγγγ 1 ZZ cc sinh γγγγ = tanh γγγγ 2 ZZ cc The equivalent ππ circuit for long transmission lines (>250 km) is
47 Long-Line vs. Medium-Line Models 47 We can compare this equivalent ππ circuit with the nominal ππ circuit used for medium-length lines, where ZZ = zzzz and YY 2 = yy ll 2 Rewriting (18) using the definition for characteristic impedance, ZZ = zz yy sinh γγγγ = zzzz zz ZZ = zzzz ZZ = ZZ sinh γγγγ zzzz ll sinh γγγγ γγγγ yy sinh γγγγ We see that the series impedance of the long-line model is equal to that of the medium-line model, multiplied by a correction factor zzzz (20)
48 Long-Line vs. Medium-Line Models 48 Doing the same for the shunt admittance, we have YY γγγγ 2 2 = yy zz tanh = yyyy 2 yy zz tanh yyyy 2 γγγγ 2 YY = yyyy 2 2 tanh γγγγ 2 zzzz ll 2 YY 2 = YY 2 tanh γγγγ 2 γγγγ 2 Again, we see a similar correction factor relating the admittance, YY, of the lumped, nominal ππ circuit to the admittance of the distributed, equivalent ππ circuit, YYY
49 49 Lossless Lines
50 Lossless Lines 50 Transmission line models can be simplified significantly if we neglect loss Sacrifice accuracy for the sake of simplicity Series resistance, RR, and shunt conductance, GG, are the model parameters accounting for loss Let RR 0 and GG 0 (we ve already assumed GG = 0) Propagation constant for a lossless line is γγ = jjjj The attenuation constant is now zero, αα 0 γγ = zzzz = jjjjjj jjjjjj = jjjj LLLL = jjjj ββ = ωω LLLL
51 Lossless Lines ABCD Parameters 51 Using the propagation constant for a lossless line, the distributed model chain parameters become AA xx = DD xx = cosh jjjjjj = eejjjjjj + ee jjjjjj AA xx = DD xx = cos ββββ BB xx = ZZ cc sinh jjjjjj = ZZ cc ee jjjjjj ee jjjjjj BB xx = jjzz cc sin ββββ CC xx = 1 ZZ cc sinh jjjjjj = 1 ZZ cc ee jjjjjj ee jjjjjj sin ββββ CC xx = jj ZZ cc 2 2 2
52 Lossless Lines ABCD Parameters 52 Chain parameters at a distance xx from the end of a lossless line are AAAAAAAA xx = cos ββββ sin ββββ jj ZZ cc jjzz cc sin ββββ cos ββββ And at the sending end of a line of length ll, xx ll, and we have AAAAAAAA = cos ββll sin ββll jj ZZ cc jjzz cc sin ββll cos ββll The characteristic impedance of the lossless line is called the surge impedance ZZ cc = zz yy = jjjjjj jjjjjj = LL CC
53 Equivalent ππ Circuit Lossless Line 53 For the lossless line so, γγ = jjjj ZZ = ZZ cc sinh jjjjjj = jj LL sin ββββ = jjxx CC and, YY 2 = tanh ZZ cc jjjjjj 2 = jj tan ββββ 2 ZZ cc
54 Wavelength 54 The voltage along the lossless line is VV xx = AA xx VV RR + BB xx II RR VV xx = cos ββββ VV RR + jjzz cc sin ββββ II RR A wavelength, λλ, is the distance required for a phase shift of 360 along the line There is a 360 phase shift when xx = λλ and ββββ = 2ππ The wavelength is λλ = 2ππ ββ = 2ππ ωω LLLL = 1 ff LLLL = νν ff where νν = 1/ LLLL is the propagation velocity along the line
55 Wavelength 55 For overhead transmission lines, νν cc mm/ss That is, electrical waves propagate along the line at roughly the speed of light At 60 Hz, the wavelength is λλ = νν ff = = 5000 kkkk This is approximately the distance across the U.S. Most transmission lines are significantly shorter than a wavelength
56 Surge Impedance Loading (SIL) 56 Surge impedance loading (SIL) The power delivered by a transmission line to a resistive load whose impedance is equal to the surge impedance, ZZ cc, of that transmission line At SIL, the load current is II RR = VV RR ZZ cc The voltage along the line is VV xx = cos ββββ VV RR + jjzz cc sin ββββ II RR VV xx = cos ββββ VV RR + jjzz cc sin ββββ VV RR ZZ cc VV xx = VV RR cos ββββ + jj sin ββββ VV xx = VV RR ββββ Note that at SIL, the magnitude of the voltage is constant along the line A flat voltage profile
57 Surge Impedance Loading (SIL) 57 At SIL, the current along the line is given by II xx = jj sin ββββ ZZ cc VV RR + cos ββββ VV RR ZZ cc II xx = VV RR ZZ cc cos ββββ + jj sin ββββ II xx = VV RR ZZ cc ββββ The complex power along the line is At SIL SS xx = VV xx II xx = VV RR ββββ SS xx = VV RR 2 ZZ cc = PP xx + jjjj xx Power flow is independent of position along the line Reactive power is zero VV RR ZZ cc ββββ
58 Surge Impedance Loading (SIL) 58 Surge impedance loading is typically defined in terms of a transmission line s rated voltage SSSSSS = VV 2 rrrrrrrrrr ZZ cc At SIL, we ve seen that the voltage profile along a transmission line is flat At no load, II RR = 0, and the voltage is given by The source voltage is VV xx = cos ββββ VV RRRRRR VV SS = cos ββββ VV RRRRRR So the receiving-end voltage in terms of the sending-end voltage is VV RRRRRR = VV SS cos ββββ
59 Surge Impedance Loading (SIL) 59 The no-load receiving-end voltage is VV RRRRRR = VV SS cos ββββ As long as ββββ ππ/2, i.e. ll λλ/4, Voltage will increase along the length of the line No-load receiving-end voltage is greater than the sending-end voltage Voltage regulation worsens with increasing line length source: Glover, Sarma, Overbye
60 Real Power vs. Voltage Angle 60 Assume a voltage angle between the sending and receiving ends of a lossless line of δδ VV RR = VV RR 0 and VV SS = VV SS δδ Using the equivalent ππ network for the lossless line, we can determine the receiving-end current Applying KCL at the receiving end II RR = VV SS VV RR jjxx jj BB 2 VV RR II RR = VV SS δδ VV RR 0 jjxx jj BB 2 VV RR 0
61 Real Power vs. Voltage Angle 61 The complex power at the load is SS RR = VV RR II RR = VV RRVV SS δδ VV RR 2 SS RR = jj VV RRVV SS δδ XX jjxx jj VV 2 RR BB + jj XX 2 VV RR 2 + jj BB 2 VV RR 2 SS RR = jj VV RRVV SS XX cos δδ + jj sin δδ jj VV 2 RR BB + jj XX 2 VV RR 2 SS RR = VV RRVV SS XX The real power delivered is sin δδ + jj VV RRVV SS XX cos δδ VV 2 RR XX + BB 2 VV RR 2 PP RR = PP SS = Re SS RR = VV RRVV SS XX sin δδ
62 Power Flow Lossless Lines 62 The delivered power is a function of the voltage phase shift along the line, δδ PP RR = VV RRVV SS XX sin δδ For the lossless line the series reactance is XX = ZZ cc sin(ββββ) so, PP RR = VV RRVV SS VV sin δδ = ZZ cc sin(ββββ) RR VV SS ZZ cc sin 2ππππ λλ sin δδ
63 Power Flow Lossless Lines 63 Converting VV RR and VV SS to per unit PP RR = VV RR VV rrrrrrrrrr VV SS VV rrrrrrrrrr 2 VV rrrrrrrrrr ZZ cc sin 2ππππ λλ sin δδ PP RR = VV RR,pppp VV SS,pppp 2 VV rrrrrrrrrr ZZ cc sin δδ sin 2ππππ λλ The term in parentheses is SIL, so PP RR = VV RR,pppp VV SS,pppp SSSSSS sin δδ sin 2ππππ λλ This provides a relationship between: Power delivered over a transmission line Voltage drop along the line Power angle
64 Maximum Power Flow Lossless Lines 64 PP RR = VV RR VV SS ZZ cc sin 2ππππ λλ sin δδ = VV RR,pppp VV SS,pppp SSSSSS sin δδ sin 2ππππ λλ The delivered power is a function of the voltage phase shift along the line Maximum power occurs when δδ = 90 PP mmmmmm = VV RR VV SS ZZ cc sin 2ππππ λλ = VV RR,ppppVV SS,pppp SSSSSS sin 2ππππ λλ The steady-state stability limit of a line
65 Steady-State Stability Limit 65 PP mmmmmm = VV RR VV SS ZZ cc sin 2ππππ λλ = VV RR,ppppVV SS,pppp SSSSSS sin 2ππππ λλ This maximum power is the steady-state stability limit of a transmission line Loads exceeding this limit will result in a loss of synchronism at the receiving end Synchronous machines at the sending and receiving ends will fall out of synchronization Steady-state stability limit proportional to Inverse of line length Square of the line voltage
66 Transmission Line Loadability 66 Three primary factors limit power flow over transmission lines: Phase shift Voltage drop Thermal limit Relevant limit depends on line length Phase shift: Proportional to line length and power flow Phase shift places a stability limit on power flow Exceeding PP mmmmmm (δδ = 90 ) results in loss of synchronism For satisfactory transient stability, typically δδ Stability limits the loadability of long transmission lines (>150 mi)
67 Transmission Line Loadability 67 Voltage drop: Voltage drop along a line is also proportional to line length and power flow Typically, voltage drop limited to 5% 10% Voltage drop limits power flow on medium-length lines (50mi 150 mi) Thermal limits As power flow increases, line temperature increases As temperature increases, lines sag and loose tensile strength A line s thermal limit is independent of line length Thermal limits dominate for short lines (<50 mi)
68 Transmission Line Loadability 68 Comparison of theoretical and practical loadability limits Practical limit assumes: VV RR /VV ss 0.95 δδ source: Glover, Sarma, Overbye
69 Practical Line Loadability Example 69 Determine how much power that can be transmitted over a 400 km, 500 kv transmission line, given the following: Voltage drop along the line limited to 10% Power angle limited to δδ mmmmmm = 30 The characteristic impedance of the line is ZZ cc = 280Ω Assume VV SS,pppp = 1.0 pp. uu. Power delivered to the receiving end of the line is PP RR = VV RR,pppp VV SS,pppp SSSSSS sin δδ sin 2ππππ λλ PP RR = SSSSSS sin sin 30 2ππ 400 kkkk 5000 kkkk
70 Practical Line Loadability Example 70 In terms of SIL, the power the line can deliver is PP RR = SSSSSS Surge impedance loading for the line is so, SSSSSS = VV 2 rrrrrrrrrr = ZZ cc 500 kkkk Ω PP RR = MMMM = MMMM PP RR = 834 MMMM
71 71 Reactive Compensation
72 Reactive Compensation 72 Voltage profile and loadability of a transmission line depend on relative line and load impedances By varying line impedance, we can affect voltage regulation and line loadability Add shunt or series reactance to the line reactive compensation Types of reactive compensation Shunt reactors (inductors) Absorb reactive power Reduce receiving-end voltage under light load Must be removed under higher-load conditions Shunt capacitors Supply reactive power Increase receiving-end voltage at full load Removed under light-load conditions
73 Reactive Compensation 73 Types of reactive compensation (cont d) Series capacitors Reduce series line impedance Reduce line voltage drops Increase steady-state stability limit Static VAR compensators (SVCs) Thyristor-controlled shunt reactors and capacitors Automatically adjust compensation depending on load
74 Reactive Compensation 74 Amount of reactive compensation is typically expressed as a percentage of line impedance For example, the circuit above shows a transmission line with NNNNN shunt reactive compensation
75 Reactive Compensation Example Consider a 300 km, 765 kv, three-phase transmission line with the following chain parameters: AA = BB = ZZ = Shunt reactors, switched in during light-load conditions only, provide 75% compensation Full-load current is 1.9 ka at 730 kv with unity power factor The sending-end voltage, VV SS, is constant Determine: %VVVV of the uncompensated line %VVVV of the compensated line
76 Reactive Compensation Example Full-load, line-to-neutral, receiving-end voltage, using it as the 0 phase reference: VV RRRRRR = kkkk = kkkk Use chain parameters to determine the sending-end voltage, VV SS VV SS = AAVV RRRRRR + BBII RRRRRR VV SS = ( kkkk) Ω VV SS = kkkk The no-load, line-to-neutral, receiving-end voltage is VV RRRRRR = VV SS AA = kkkk = kkkk Percent voltage regulation for the uncompensated line is kkkk %VVVV = VV RRRRRR VV RRRRRR VV RRRRRR 100% = kkkk kkkk kkkk 100% %VVVV = 12.7%
77 Reactive Compensation Example For the compensated line, we need to calculate new chain parameters Shunt admittance of the uncompensated line can be determined from the known chain parameters where So, AA = = 1 + YY ZZ ZZ = BB = Ω YY = AA 1 2 ZZ = Ω YY = SS YY = jjj SS 2
78 Reactive Compensation Example After adding compensation, the equivalent shunt susceptance decreases by 75% YY eeee = jjj SS 0.25 YY eeee = jjjjj 10 6 SS Use YY eeee to calculate the AA parameter for the compensated line AA eeee = 1 + YY eeeezz 2 = Note that shunt reactive compensation does not affect the series impedance, ZZZ, and therefor does not affect BB
79 Reactive Compensation Example The no-load receiving-end voltage for the compensated line: VV RRRRRR = VV SS AA eeee = kkkk VV RRRRLL = kkkk Percent voltage regulation for the compensated line is %VVVV = VV RRRRRR VV RRRRRR 100% VV RRRRRR %VVVV = kkkk kkkk kkkk 100% %VVVV = 6.8% Reactive compensation has improved voltage regulation from 12.7% to 6.8%
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